Lemma. a Hurwitz unit of order \(3\) [fgap-0003]

Let \[ a=i+j+k,\qquad \omega =\frac {-1+i+j+k}{2}=\frac {-1+a}{2}. \] Then \(\omega \) is a Hurwitz unit of order \(3\). This choice appears in [voight2021quaternion, p. 165].

The mixed terms in \(a^2\) cancel in pairs: \[ \begin {aligned} a^2 &=i^2+j^2+k^2+(ij+ji)+(jk+kj)+(ki+ik)\\ &=-3. \end {aligned} \] It follows that \[ \begin {aligned} \omega ^2 &=\frac {(-1+a)^2}{4}\\ &=\frac {1-2a+a^2}{4}\\ &=\frac {-1-a}{2}. \end {aligned} \] Hence \(\omega ^2+\omega +1=0\), and multiplication by \(\omega -1\) gives \[ \omega ^3=1. \] Thus \(\omega \) is invertible, has inverse \(\omega ^{-1}=\omega ^2=(-1-i-j-k)/2\), and has order \(3\) because \(\omega \ne 1\).